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I want to find some subset of R^2 which its intersection with every vertical line is measure zero if we see it as a subset of R and it is not measure zero in R^2?

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Why?${}{}{}{}{}$ – Asaf Karagila Mar 14 '14 at 11:27
because of some reasons in integrability of multivariable functions – alich Mar 14 '14 at 11:29
Wouldn't this contradict Fubini's theorem? Of course, assuming that the set is measurable in the first place. – Alex Degtyarev Mar 14 '14 at 11:31
Why does It contradict Fubini theorem? – alich Mar 14 '14 at 11:35
@alich: Because such a set $A\subseteq\mathbb{R}^2$ would satisfy $0\neq \lambda^2(A) = \int_{\mathbb{R}^2} \chi_A(x,y) d\lambda^2(x,y) = \int_\mathbb{R} \int_\mathbb{R} \chi_A(x,y) d\lambda^1(x) d\lambda^1(y) = \int_\mathbb{R} 0 d\lambda^1(y) = 0$. (where $\lambda^{1,2}$ denotes the 1-dimensional and 2-dimensional lebesgue meausure respectively) – Johannes Hahn Mar 14 '14 at 11:59
up vote 9 down vote accepted

It is a well known result of Sierpinski that there exists non-measurable subsets of $\mathbb{R}^{2}$ which intersect each line in at most two points. Furthermore, there exists a real-valued function whose graph is a non-measurable subset of $\mathbb{R}^{2}$.

  1. W. Sierpi´nski: Sur un probl`eme concernant les ensembles mesurables superficiellement. Fund. Math. 1 (1920), 112–115.

  2. Gelbaum, Bernard R., and John M. H. Olmsted. Counterexamples in Analysis. San Francisco: Holden-Day, 1964. (p. 142-145)

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